If the sums of n terms of two arithmetic progressions are in the ratio 3n + 5/5n - 7 , then their nth terms are in the ratio

Question

KollegeApply · Accepted Answer

In the given problem, the ratio of the sum of n terms of two A.P’s is given by the expression, `(S_n)/(S'_n) = (3n + 5)/(5n+ 7)` ....(1) We need to find the ratio of their nth terms. Here we use the following formula for the sum of n terms of an A.P., S_n = n/2 [ 2a + ( n - 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms So, S_n = n/2 [ 2a + ( n - 1)d] Where, a and d are the first term and the common difference of the first A.P. Similarly, `S'_n = n/2 [ 2a' + ( n - 1)d']` Where, a’ and d’ are the first term and the common difference of the first A.P. So, `(S_n)/(S'_n) = (n/2[2a + (n-1)d])/(n/2[2a'+(n-1)d']) = ([2a + (n-1)d])/([2a'+(n-1)d'])` ............(2) Equating (1) and (2), we get, `= ( [2a + (n-1)d])/( [2a'+(n-1)d'])=(3n + 5)/(5n + 7)` Now, to find the ratio of the nth term, we replace n by2n - 1 . We get, `= ( [2a + (n-1-1)d])/( [2a'+(n-1-1 )d'])= (3(2n - 1) + 5)/(5(2n -1) + 7) ( 2a + (2n-2)d)/( 2a'+(2n-2)d')= (6n - 3 + 5 )/(10n - 5 + 7) ( 2a + 2(n-1)d)/( 2a'+2(n-1)d')=(6n +2)/(10n + 2) ( a + (n-1)d)/( a'+(n-1)d') = (3n + 1)/(5n + 1)` As we know, an = a + (n - 1)d Therefore, we get, `a_n /(a'_n) = (3n + 1) /(5n +1)`

AChoose Your Answer